Electromagnetic Oscillations and Alternating Current

1. Electromagnetic Oscillations andAlternating Current Chapter 33

2. Oscillations in an LC Circuit We will discover that charge sloshes back and forth. As this happens the current goes one way then the other. Analogy: a block moving on a spring. Here total energy (kinetic + potential) is constant. For the LC circuit total energy (electric + magnetic) is constant.

3. i i + + + time i i LC Circuit i i + + +

4. Analyzing an LC Circuit Total energy in the circuit: No change in energy Differentiate :

5. Analyzing an LC Circuit Total energy in the circuit: No change in energy Differentiate :

6. Analyzing an LC Circuit Total energy in the circuit: No change in energy Differentiate :

7. Analyzing an LC Circuit Total energy in the circuit: No change in energy Differentiate : The charge sloshes back and forth with frequency w = (LC)-1/2

8. Analyzing an LC Circuit Current Current is maximum when charge is zero, and vice versa. Energy:

9. RLC Circuit: Damped Oscillations R C L The change here is that energy is dissipated in the resistor: A similar analysis gives current and charge that continue to oscillate but with amplitudes that decay exponentially:

10. V(t) Vp wt p 2p fv -Vp An “AC” circuit is one in which the driving voltage and hence the current are sinusoidal in time. Alternating Current Circuits V = VP sin (wt - fv ) I = IP sin (wt - fI ) wis theangular frequency (angular speed) [radians per second]. Sometimes instead ofwwe use thefrequency f [cycles per second] Frequency f [cycles per second, or Hertz (Hz)]w= 2p f

11. V(t) Vp wt p 2p fv -Vp V = VP sin (wt - fv ) I = IP sin (wt - fI ) Alternating Current Circuits I(t) Ip Irms Vrms t fI/w -Ip Vp and Ip are the peak current and voltage. We also use the “root-mean-square” values: Vrms = Vp / and Irms=Ip / fv andfI are called phase differences (these determine when V and I are zero). Usually we’re free to set fv=0 (but not fI).

12. Example: household voltage In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0.

13. Example: household voltage In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0. This 120 V is the RMS amplitude: so Vp=Vrms = 170 V.

14. Example: household voltage In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0. This 120 V is the RMS amplitude: so Vp=Vrms = 170 V. This 60 Hz is the frequency f: so w=2p f=377 s -1.

15. Example: household voltage In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0. This 120 V is the RMS amplitude: so Vp=Vrms = 170 V. This 60 Hz is the frequency f: so w=2p f=377 s -1. So V(t) = 170 sin(377t + fv). Choose fv=0 so that V(t)=0 at t=0: V(t) = 170 sin(377t).

16. R V ~ Resistors in AC Circuits EMF (and also voltage across resistor): V = VP sin (wt) Hence by Ohm’s law, I=V/R: I = (VP /R) sin(wt) = IP sin(wt) (with IP=VP/R) V I V and I “In-phase” p wt 2p

17. C Start from: q = C V [V=Vpsin(wt)] Take derivative: dq/dt = C dV/dt So I = C dV/dt = C VPw cos (wt) I = C w VP sin (wt + p/2) Capacitors in AC Circuits V ~ This looks like IP=VP/R for a resistor (except for the phase change). So we call Xc = 1/(wC) the Capacitive Reactance V I wt p 2p The reactance is sort of like resistance in that IP=VP/Xc. Also, the current leads the voltage by 90o (phase difference). V and I “out of phase” by 90º. I leads V by 90º.

18. C V ~ Capacitor Example A 100 nF capacitor is connected to an AC supply of peak voltage 170V and frequency 60 Hz. What is the peak current? What is the phase of the current? What is the dissipated power?

19. Inductors in AC Circuits V = VP sin (wt) Loop law: V +VL= 0 where VL = -L dI/dt Hence: dI/dt = (VP/L) sin(wt). Integrate: I = - (VP / Lw) cos (wt) or I = [VP /(wL)] sin (wt - p/2) ~ L V Again this looks like IP=VP/R for a resistor (except for the phase change). So we call XL = w L the Inductive Reactance I p wt 2p Here the current lags the voltage by 90o. V and I “out of phase” by 90º. I lags V by 90º.

20. V ~ L Inductor Example A 10 mH inductor is connected to an AC supply of peak voltage 10V and frequency 50 kHz. What is the peak current? What is the phase of the current? What is the dissipated power?

22. Phasor Diagrams A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. The “y component” is the actual voltage or current. Resistor Vp Ip w t

23. Phasor Diagrams A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. The “y component” is the actual voltage or current. Resistor Capacitor Vp Ip Ip w t w t Vp

24. Phasor Diagrams A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. The “y component” is the actual voltage or current. Resistor Capacitor Inductor Vp Vp Ip Ip Ip w t w t w t Vp